Problem: Graph this system of equations and solve. $-2x-4y = -8$ $3x-4y = 12$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $-2x-4y = -8$ , to slope-intercept form. $y = -\dfrac{1}{2} x + 2$ The y-intercept for the first equation is $2$ , so the first line must pass through the point $(0, 2)$ The slope for the first equation is $-\dfrac{1}{2}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $2$ positions to the right. $2$ positions to the right. Graph the blue line so it passes through $(0, 2)$ and $(2, 1)$ Convert the second equation, $3x-4y = 12$ , to slope-intercept form. $y = \dfrac{3}{4} x - 3$ The y-intercept for the second equation is $-3$ , so the second line must pass through the point $(0, -3)$ The slope for the second equation is $\dfrac{3}{4}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up You must also move $4$ positions to the right. $4$ positions to the right. $3$ positions up from $(0, -3)$ is $(4, 0)$ Graph the green line so it passes through $(0, -3)$ and $(4, 0)$ The solution is the point where the two lines intersect. The lines intersect at $(4, 0)$.